MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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### Statistical Hypothesis Testing: Mean Tuna Consumption A nutritionist claims that the mean tuna consumption by a person is 3.3 pounds per year. A sample of 60 people shows that the mean tuna consumption by a person is 3.1 pounds per year. Assume the population standard deviation is 1.24 pounds. At \( \alpha = 0.06 \), can you reject the claim? #### Step-by-Step Solution 1. **Calculate the Test Statistic:** \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \] Given: - \( \bar{x} = 3.1 \) - \( \mu = 3.3 \) - \( \sigma = 1.24 \) - \( n = 60 \) \[ z = \frac{3.1 - 3.3}{1.24 / \sqrt{60}} \approx -1.25 \quad \text{(Rounded to two decimal places as needed.)} \] 2. **Find the P-value:** Using the standard normal distribution table (Z-table), we find the area to the left of \( z = -1.25 \). \[ \text{P-value} = 0.211 \quad \text{(Rounded to three decimal places as needed.)} \] 3. **Decision Rule:** Compare the P-value with the significance level \( \alpha = 0.06 \). - If \( \text{P-value} \leq \alpha \), reject the null hypothesis \( H_0 \). - If \( \text{P-value} > \alpha \), fail to reject the null hypothesis \( H_0 \). Here, \( \text{P-value} = 0.211 \) which is greater than \( \alpha = 0.06 \). 4. **Conclusion:** - Fail to reject \( H_0 \). There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds. ### Multiple Choice Decision #### Decide whether to reject or fail to reject the null hypothesis: - **A. Fail to reject \( H_0 \). There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds
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Transcribed Image Text:### Statistical Hypothesis Testing: Mean Tuna Consumption A nutritionist claims that the mean tuna consumption by a person is 3.3 pounds per year. A sample of 60 people shows that the mean tuna consumption by a person is 3.1 pounds per year. Assume the population standard deviation is 1.24 pounds. At \( \alpha = 0.06 \), can you reject the claim? #### Step-by-Step Solution 1. **Calculate the Test Statistic:** \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \] Given: - \( \bar{x} = 3.1 \) - \( \mu = 3.3 \) - \( \sigma = 1.24 \) - \( n = 60 \) \[ z = \frac{3.1 - 3.3}{1.24 / \sqrt{60}} \approx -1.25 \quad \text{(Rounded to two decimal places as needed.)} \] 2. **Find the P-value:** Using the standard normal distribution table (Z-table), we find the area to the left of \( z = -1.25 \). \[ \text{P-value} = 0.211 \quad \text{(Rounded to three decimal places as needed.)} \] 3. **Decision Rule:** Compare the P-value with the significance level \( \alpha = 0.06 \). - If \( \text{P-value} \leq \alpha \), reject the null hypothesis \( H_0 \). - If \( \text{P-value} > \alpha \), fail to reject the null hypothesis \( H_0 \). Here, \( \text{P-value} = 0.211 \) which is greater than \( \alpha = 0.06 \). 4. **Conclusion:** - Fail to reject \( H_0 \). There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds. ### Multiple Choice Decision #### Decide whether to reject or fail to reject the null hypothesis: - **A. Fail to reject \( H_0 \). There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds
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